Prove Theorem 10.1: Let \(T: V \rightarrow V\) be a linear operator whose
characteristic polynomial factors into linear polynomials. Then \(V\) has a
basis in which \(T\) is represented by a triangular matrix.
The proof is by induction on the dimension of \(V\). If \(\operatorname{dim}
V=1\), then every matrix representation of \(T\) is a \(1 \times 1\) matrix, which
is triangular.
Now suppose \(\operatorname{dim} V=n>1\) and that the theorem holds for spaces
of dimension less than \(n .\) Because the characteristic polynomial of \(T\)
factors into linear polynomials, \(T\) has at least one eigenvalue and so at
least one nonzero eigenvector \(v\), say \(T(v)=a_{11} v .\) Let \(W\) be the one-
dimensional subspace spanned by \(v\). Set \(\bar{V}=V / W .\) Then (Problem
10.26) \(\operatorname{dim} \bar{V}=\operatorname{dim} V-\operatorname{dim}
W=n-1 .\) Note also that \(W\) is invariant under \(T .\) By Theorem \(10.16, T\)
induces a linear operator \(\bar{T}\) on \(\bar{V}\) whose minimal polynomial
divides the minimal polynomial of \(T\). Because the characteristic polynomial
of \(T\) is a product of linear polynomials, so is its minimal polynomial, and
hence, so are the minimal and characteristic polynomials of \(\bar{T}\). Thus,
\(\bar{V}\) and \(\bar{T}\) satisfy the hypothesis of the theorem. Hence, by
induction, there exists a basis \(\left\\{\bar{v}_{2}, \ldots,
\bar{v}_{n}\right\\}\) of \(\bar{V}\) such that
$$
\begin{aligned}
&\bar{T}\left(\bar{v}_{2}\right)=a_{22} \bar{v}_{2} \\
&\bar{T}\left(\bar{v}_{3}\right)=a_{32} \bar{v}_{2}+a_{33} \bar{v}_{3} \\
&\cdots \ldots \ldots \cdots \\
&\bar{T}\left(\bar{v}_{n}\right)=a_{n 2} \bar{v}_{n}+a_{n 3}
\bar{v}_{3}+\cdots+a_{n n} \bar{v}_{n}
\end{aligned}
$$Now let \(v_{2}, \ldots, v_{n}\) be elements of \(V\) that belong to the cosets
\(v_{2}, \ldots, v_{n}\), respectively. Then \(\left\\{v, v_{2}, \ldots,
v_{n}\right\\}\) is a basis of \(V\) (Problem 10.26). Because
\(\bar{T}\left(v_{2}\right)=a_{22} \bar{v}_{2}\), we have
$$
\bar{T}\left(\bar{v}_{2}\right)-a_{22} \bar{v}_{22}=0, \quad \text { and so }
\quad T\left(v_{2}\right)-a_{22} v_{2} \in W
$$
But \(W\) is spanned by \(v\); hence, \(T\left(v_{2}\right)-a_{22} v_{2}\) is a
multiple of \(v\), say,
$$
T\left(v_{2}\right)-a_{22} v_{2}=a_{21} v, \quad \text { and so } \quad
T\left(v_{2}\right)=a_{21} v+a_{22} v_{2}
$$
Similarly, for \(i=3, \ldots, n\)
$$
T\left(v_{i}\right)-a_{i 2} v_{2}-a_{i 3} v_{3}-\cdots-a_{i i} v_{i} \in W,
\quad \text { and so } \quad T\left(v_{i}\right)=a_{i 1} v+a_{i 2}
v_{2}+\cdots+a_{i i} v_{i}
$$
Thus,
$$
\begin{aligned}
&T(v)=a_{11} v \\
&T\left(v_{2}\right)=a_{21} v+a_{22} v_{2} \\
&\ldots \ldots \ldots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
&T\left(v_{n}\right)=a_{n 1} v+a_{n 2} v_{2}+\cdots+a_{n n} v_{n}
\end{aligned}
$$
and hence the matrix of \(T\) in this basis is triangular.
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